What is the DFT of DFT of discrete signal











up vote
1
down vote

favorite












What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?






discrete-signals fourier-transform dft







share|improve this question


























    up vote
    1
    down vote

    favorite












    What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?






    discrete-signals fourier-transform dft







    share|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?






      discrete-signals fourier-transform dft







      share|improve this question













      What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?






      discrete-signals fourier-transform dft




      discrete-signals fourier-transform dft






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 8 hours ago









      Mert Ege

      133




      133






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote













          let



          $$begin{align}
          X[k] &= mathcal{DFT} Big{ x[n] Big} \
          &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
          end{align} $$



          and



          $$ y[n] triangleq X[n] $$



          (note the substitution of $n$ in for $k$.) then



          $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



          then, if the DFT is defined the most common way (as above):



          $$ Y[n] = N cdot x[-n] $$



          where periodicity is implied: $x[n+N]=x[n]$ for all $n$.






          share|improve this answer




























            up vote
            1
            down vote













            Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



            Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).






            share|improve this answer























            • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
              – robert bristow-johnson
              7 hours ago











            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "295"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdsp.stackexchange.com%2fquestions%2f53959%2fwhat-is-the-dft-of-dft-of-discrete-signal%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote













            let



            $$begin{align}
            X[k] &= mathcal{DFT} Big{ x[n] Big} \
            &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
            end{align} $$



            and



            $$ y[n] triangleq X[n] $$



            (note the substitution of $n$ in for $k$.) then



            $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



            then, if the DFT is defined the most common way (as above):



            $$ Y[n] = N cdot x[-n] $$



            where periodicity is implied: $x[n+N]=x[n]$ for all $n$.






            share|improve this answer

























              up vote
              4
              down vote













              let



              $$begin{align}
              X[k] &= mathcal{DFT} Big{ x[n] Big} \
              &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
              end{align} $$



              and



              $$ y[n] triangleq X[n] $$



              (note the substitution of $n$ in for $k$.) then



              $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



              then, if the DFT is defined the most common way (as above):



              $$ Y[n] = N cdot x[-n] $$



              where periodicity is implied: $x[n+N]=x[n]$ for all $n$.






              share|improve this answer























                up vote
                4
                down vote










                up vote
                4
                down vote









                let



                $$begin{align}
                X[k] &= mathcal{DFT} Big{ x[n] Big} \
                &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
                end{align} $$



                and



                $$ y[n] triangleq X[n] $$



                (note the substitution of $n$ in for $k$.) then



                $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



                then, if the DFT is defined the most common way (as above):



                $$ Y[n] = N cdot x[-n] $$



                where periodicity is implied: $x[n+N]=x[n]$ for all $n$.






                share|improve this answer












                let



                $$begin{align}
                X[k] &= mathcal{DFT} Big{ x[n] Big} \
                &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
                end{align} $$



                and



                $$ y[n] triangleq X[n] $$



                (note the substitution of $n$ in for $k$.) then



                $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



                then, if the DFT is defined the most common way (as above):



                $$ Y[n] = N cdot x[-n] $$



                where periodicity is implied: $x[n+N]=x[n]$ for all $n$.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 7 hours ago









                robert bristow-johnson

                10.4k31448




                10.4k31448






















                    up vote
                    1
                    down vote













                    Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



                    Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).






                    share|improve this answer























                    • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                      – robert bristow-johnson
                      7 hours ago















                    up vote
                    1
                    down vote













                    Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



                    Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).






                    share|improve this answer























                    • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                      – robert bristow-johnson
                      7 hours ago













                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



                    Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).






                    share|improve this answer














                    Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



                    Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 7 hours ago

























                    answered 7 hours ago









                    hotpaw2

                    25.5k53472




                    25.5k53472












                    • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                      – robert bristow-johnson
                      7 hours ago


















                    • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                      – robert bristow-johnson
                      7 hours ago
















                    shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                    – robert bristow-johnson
                    7 hours ago




                    shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                    – robert bristow-johnson
                    7 hours ago


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Signal Processing Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdsp.stackexchange.com%2fquestions%2f53959%2fwhat-is-the-dft-of-dft-of-discrete-signal%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Contact image not getting when fetch all contact list from iPhone by CNContact

                    Image
                    up vote 0 down vote favorite I know this question already asked but not getting solution. From this code I will get all the information from the contact but image not found when open vcf files on mac os, also not getting when share this file. I use this stackoverflow link here but It's not help full. var contacts = [CNContact]() let keys = [CNContactVCardSerialization.descriptorForRequiredKeys() ] as [Any] let request = CNContactFetchRequest(keysToFetch: keys as! [CNKeyDescriptor]) do { try self.contactStore.enumerateContacts(with: request) { (contact, stop) in // Array containing all unified contacts from everywhere contacts.append(contact) } } catch { print("unable to fetch contacts") } do { let data = try
                    Read more

                    count number of partitions of a set with n elements into k subsets

                    Image
                    up vote 6 down vote favorite 1 This program is for count number of partitions of a set with n elements into k subsets I am confusing here return k*countP(n-1, k) + countP(n-1, k-1); can some one explain what is happening here? why we are multiplying with k? NOTE->I know this is not the best way to calculate number of partitions that would be DP // A C++ program to count number of partitions // of a set with n elements into k subsets #include<iostream> using namespace std; // Returns count of different partitions of n // elements in k subsets int countP(int n, int k) { // Base cases if (n == 0 || k == 0 || k > n) return 0; if (k == 1 || k == n) return 1; // S(n+1, k) = k*S(n, k) + S(n, k-1) return k*countP(n-1, k) + countP(n-1, k-1); } // Driver program int ma
                    Read more

                    A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

                    Image
                    0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{
                    Read more